Suppose that I have a prime ideal $p$ in $\mathbb{Q}$. Then it ramifies in some extension of $\mathbb{Q}$, namely in $\mathbb{Q}(\sqrt p)$. This seems like it should be true for an arbitrary number field replacing $\mathbb{Q}$. Namely, suppose that I have a prime ideal $\mathfrak{p}$ in a number field $K$. Should there be some algebraic extension $L$ of $K$ (or even better, a quadratic extension of $K$) in which $\mathfrak{p}$ ramifies? The only criterion of a prime ramifying (which involves the prime dividing the relative discriminant of the extension) involves the assumption that $\mathcal{O}_L$ is a free $\mathcal{O}_K$-module, but this isn't true even in the case of a quadratic extension (I believe Keith Conrad has an example written up in a paper), hence my difficulty. But mostly I feel that I'm missing something trivial.
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6$\begingroup$ Just take the square root of an element of $K$ with valuation $1$ at the prime. $\endgroup$– Torsten EkedahlCommented Oct 12, 2011 at 15:15
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6$\begingroup$ Take an element $f$ of $O_K$ which generates $\mathfrak p$ in the local ring $O_{K, \mathfrak p}$ and consider the extension $L=K(\sqrt{f})$. It is quadratic and ramifies at $\mathfrak p$ (look at the ramification index). The ramification is a local phenomena and doesn't have anything to do with the freeness of $O_L$ over $O_K$. $\endgroup$– Qing LiuCommented Oct 12, 2011 at 15:19
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$\begingroup$ Thanks, that makes sense. But why would one generally require the freeness when they state the ramification criterion? For example, Milne's Algebraic Number Theory (Thm 3.35) asserts this condition. $\endgroup$– user3860Commented Oct 12, 2011 at 15:21
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3$\begingroup$ J, you are referring in your parenthetical remark to www.math.uconn.edu/~kconrad/blurbs/gradnumthy/notfree.pdf, but this issue isn't relevant. A prime in a number field ramifies in a finite extension iff it divides the discriminant ideal of the extension, and there is no need for an initial hypothesis that the top ring of integers is a free module over the bottom ring of integers. (However, in the proof of the theorem one may localize at the prime of interest and the localized ring of integers becomes a PID, so leading to a free module as a technical convenience, not a hypothesis.) $\endgroup$– KConradCommented Oct 12, 2011 at 17:47
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2$\begingroup$ J, you are missing something trivial: see Milne's Remark 3.39(b), where he indicates how to remove the freeness assumption by using localization. $\endgroup$– KConradCommented Oct 12, 2011 at 17:52
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Why not try to "solve a harder problem", and produce Eisenstein polynomials of whatever degree you want, over any number field? These give totally ramified extensions.
answered Oct 12, 2011 at 20:52